Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x^2 - 19x + 90}{-2x^2 + 8x + 120} \div \dfrac{x + 8}{4x + 24} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{x^2 - 19x + 90}{-2x^2 + 8x + 120} \times \dfrac{4x + 24}{x + 8} $ First factor out any common factors. $q = \dfrac{x^2 - 19x + 90}{-2(x^2 - 4x - 60)} \times \dfrac{4(x + 6)}{x + 8} $ Then factor the quadratic expressions. $q = \dfrac {(x - 10)(x - 9)} {-2(x - 10)(x + 6)} \times \dfrac {4(x + 6)} {x + 8} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (x - 10)(x - 9) \times 4(x + 6)} { -2(x - 10)(x + 6) \times (x + 8)} $ $q = \dfrac {4(x - 10)(x - 9)(x + 6)} {-2(x - 10)(x + 6)(x + 8)} $ Notice that $(x - 10)$ and $(x + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {4\cancel{(x - 10)}(x - 9)(x + 6)} {-2\cancel{(x - 10)}(x + 6)(x + 8)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $q = \dfrac {4\cancel{(x - 10)}(x - 9)\cancel{(x + 6)}} {-2\cancel{(x - 10)}\cancel{(x + 6)}(x + 8)} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $q = \dfrac {4(x - 9)} {-2(x + 8)} $ $ q = \dfrac{-2(x - 9)}{x + 8}; x \neq 10; x \neq -6 $